Find $\dfrac{d}{dx}\left(\ln(x)e^x\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $e^x\left(\dfrac1x+\ln(x)\right)$ (Choice B) B $\dfrac{e^x}{x}$ (Choice C) C $\dfrac1x+e^x$ (Choice D) D $\dfrac{e^x}{x}+\ln(x)$
Explanation: $\ln(x)e^x$ is the product of two, more basic, expressions: $\ln(x)$ and $e^x$. Therefore, the derivative of the expression can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\ln(x)e^x\right) \\\\ &=\dfrac{d}{dx}\left(\ln(x)\right)e^x+\ln(x)\dfrac{d}{dx}(e^x)&&\gray{\text{The product rule}} \\\\ &=\dfrac1x\cdot e^x+\ln(x)\cdot e^x&&\gray{\text{Differentiate }\ln(x)\text{ and }e^x} \\\\ &=e^x\left(\dfrac1x+\ln(x)\right)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(\ln(x)e^x\right)=e^x\left(\dfrac1x+\ln(x)\right)$